FINITELY ADDITIVE, MODULAR, AND PROBABILITY FUNCTIONS …

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FINITELY ADDITIVE, MODULAR, AND PROBABILITY

FUNCTIONS ON PRE-SEMIRINGS

PEYMAN NASEHPOUR AND AMIR HOSSEIN PARVARDI

Abstract. In this paper, we define finitely additive, probability and mod-ular functions over semiring-like structures. We investigate finitely additivefunctions with the help of complemented elements of a semiring. We also gen-eralize some classical results in probability theory such as the Law of TotalProbability, Bayes’ Theorem, the Equality of Parallel Systems, and Poincare’sInclusion-Exclusion Theorem. While we prove that modular functions over acouple of known semirings are almost constant, we show it is possible to de-fine many different modular functions over some semirings such as bottleneckalgebras and the semiring (Id(D),+, ·), where D is a Dedekind domain.

0. Introduction

Semirings and other semiring-like algebraic structures such as pre-semirings,hemirings and near-rings have many applications in engineering, especially in com-puter science and of course in applied mathematics [12], [14], [18] and [25]. Finitelyadditive and modular functions appear in measure theory [4], probability theory[9], lattice and Boolean algebra theory [15], module theory [2], and a couple ofother branches of mathematics as we will mention in different parts of the paper.Since semiring-like algebraic structures are interesting generalizations of distribu-tive lattices and rings, it is quite natural to ask if finitely additive and modularfunctions can be defined and investigated in “pre-semiring theory”. On the otherhand, probability functions, as special cases for finitely additive functions, play animportant role in probability theory. Therefore it seems quite interesting to see ifthe classical results of probability theory can be stated and proved in this contextas well.

In the present paper, we define and investigate finitely additive, modular, andprobability functions over semirings and semiring-like algebraic structures. Sincedifferent authors give different definitions for semirings and similar algebraic struc-tures, it is crucial to clarify what we mean by these structures. In this paper, bya “pre-semiring”, we understand an algebraic structure, consisting of a nonemptyset S with two operations of addition and multiplication such that the followingconditions are satisfied:

(1) (S,+) and (S, ·) are commutative semigroups;(2) Distributive law: a · (b + c) = a · b+ a · c for all a, b, c ∈ S.

The definition of pre-semirings is borrowed from the book [14] and for more onpre-semirings, one may refer to that.

2010 Mathematics Subject Classification. 16Y60, 20M14.Key words and phrases. Semiring, Hemiring, Pre-semiring, Finitely additive functions, Mod-

ular functions, Probability functions.

1

http://arxiv.org/abs/1603.03201v2

2 PEYMAN NASEHPOUR AND AMIR HOSSEIN PARVARDI

An algebraic structure that is a pre-semiring and possesses an element that isa neutral element for its addition and an absorbing element for its multiplication,is called a hemiring. Usually, the neutral element of a hermring is denoted by 0.Any hemiring with a multiplicative identity 1 6= 0 is called a semiring. For moreon semirings and hemirings, one may refer to [12].

Let us recall that if B is a Boolean algebra, a real-valued function µ on B is calledfinitely additive, if µ(p ∨ q) = µ(p) + µ(q), whenever p and q are disjoint elementsof B [17, Chap. 15]. In Definition 1.1, we define a function f from a hemiring Sinto a commutative semigroup T to be finitely additive if st = 0 for any s, t ∈ Simplies that f(s+ t) = f(s)+f(t). The first section of the present paper is devotedto finitely additive functions. Let us recall that an element s of a semiring S issaid to be complemented, if there exists an element cs ∈ S satisfying scs = 0 ands+ cs = 1. The mentioned element cs ∈ S is called the complement of s ∈ S. Onecan easily check that if s ∈ S has a complement, then it is unique. The complementof s ∈ S, if it exists, is denoted by s⊥. Note that if s is complemented, then s⊥ isalso complemented and (s⊥)⊥ = s. Also, note that if s, t ∈ S are complemented,the symmetric difference of s and t is defined to be st = s⊥t+ st⊥ [12, Chap. 5].Finally a semiring S is called zerosumfree, if s+ t = 0 implies that s = t = 0 for alls, t ∈ S.

In the first section, we investigate finitely additive functions over complementedelements of semirings and show that if S is a zerosumfree semiring, T is a ring, andf : S −→ T is a finitely additive and normalized function (i.e., f(1) = 1), thenfor complemented elements s, t ∈ S, the following statements hold (See Proposition1.4, Proposition 1.6, and Proposition 1.7):

(1) f(t) = f(ts) + f(ts⊥).(2) f(st) + 2f(st) = f(s) + f(t), where st = s⊥t+ st⊥.(3) f(s) + f(s⊥) = 1.(4) f(s⊥t⊥) = 1− f(s)− f(t) + f(st).

Similar to the concept of independent events in probability theory, we definesome elements s1, s2, . . . , sn of a pre-semiring S to be independent, if

f(∏

x∈X

x) =∏

x∈X

f(x),

for any nonempty subset X of s1, s2, . . . , sn, where f is a function from S intoa commutative semigroup T (See Definition 1.8). After that, in Theorem 1.9,we prove that if S is a semiring, R a ring, and f : S −→ R a finitely additiveand normalized function, then the following statements for complemented elementss1, s2, . . . , sn ∈ S are equivalent:

(1) The elements s1, s2, . . . , sn ∈ S are independent,(2) The 2n sets of elements of the form t1, t2, . . . , tn with ti = si or ti = s⊥i are

independent,(3) The 2n equalities f(t1t2 · · · tn) = f(t1)f(t2) · · · f(tn) with ti = si or ti = s⊥i

hold.

In the second section of the present paper, we define probability functions oversemirings (Definition 2.1), inspired from the definition of probability functions inprobability theory [21, Chap. I]. For defining probability functions, we put an orderon the co-domains of those functions and define them as follows:

FINITELY ADDITIVE, MODULAR, AND PROBABILITY FUNCTIONS 3

Let S be a semiring and T an ordered semiring. We define a function p : S −→ Tto be a probability function, if the following properties are satisfied:

(1) p(s) ≥ 0, for any s ∈ S.(2) p(1) = 1.(3) If s, t ∈ S and st = 0, then p(s+ t) = p(s) + p(t).

In this section, we prove a couple of nice results for probability functions oversemirings similar to probability theory and generalize some of the classical resultsof probability theory such as Boole’s Inequality (Theorem 2.7), the Law of TotalProbability (Proposition 2.11), and Bayes’ Theorem (Theorem 2.14).

Modular functions appear in different branches of mathematics. A modularfunction is usually a real-valued function m over some objects that two operations“+′′ and “·′′ can apply over those objects and if s, t are of those objects, we havethe following “modular” equality:

m(s+ t) +m(st) = m(s) +m(t).

For example, if L is a lattice of finite length, then L is a modular lattice if and onlyif the function height : L −→ N0 is modular, i.e.,

height(a ∨ b) + height(a ∧ b) = height(a) + height(b),

for all a, b ∈ L [15, Corollary 376]. The third section of the present paper isdevoted to modular functions. In Definition 3.1, we define modular functions overpre-semirings similarly, and examine those functions over different pre-semiringssuch as Tropical, Max-plus, Truncation, Arctic, and B(n, i) algebras / semiringsand show that modular functions over these semirings are almost constant, i.e.,they are constant functions over their domains except a finite number of elementsin them (See the results from Proposition 3.3 to Theorem 3.13).

In different branches of mathematics, there are many interesting examples ofmodular functions that are not almost constant. We have gathered a couple ofthem in Example 3.21.

However, we find two interesting examples for nontrivial modular functions. Oneis given in Proposition 3.16 and the other is given in Theorem 3.18 and is as follows:

If (G,+) is an Abelian group and D is a Dedekind domain, a modular functionf : Id(D) −→ G can be characterized by the values of f at (0), D, and all maximalideals of D.

Finally, we show in Theorem 3.22 that Poincare’s inclusion-exclusion theorem inprobability theory holds for arbitrary modular functions over pre-semirings. Moreprecisely, we prove that if S is a multiplicatively idempotent pre-semiring, T acommutative semigroup, and m : S −→ T is a modular function, then for anyelements s1, s2, . . . , sn ∈ S, the following equality holds:

m

(

n∑

i=1

si

)

+∑

i<j

m(sisj) +∑

i<j<k<l

m(sisjsksl) + · · · =

n∑

i=1

m(si) +∑

i<j<k

m(sisjsk) + · · · (Poincare’s Formula),

where the number of multiplicative factors in the sums of both sides of the equalityis at most n.


As a result of Poincare’s formula, in Proposition 3.23, we prove that if S is amultiplicatively idempotent pre-semiring, R a ring, and f : S −→ R a modularfunction, then for any n > 1, if s1, s2, . . . , sn ∈ S are independent, then s1 + s2 +· · ·+ sn−1 and sn are also independent.

In the final part of the paper, we also prove that over complemented elementsof a semiring, finitely additive functions are modular. In fact, we show that if Sis a zerosumfree semiring such that 1 + 1 is a complemented element of S, T is acommutative semigroup, and f : S −→ T a finitely additive function, then

f |comp(S): comp(S) −→ T

is modular, where by comp(S), we mean the set of complemented elements of thesemiring S (See Theorem 3.24).

We emphasize that in this paper, the multiplication in all rings, semirings, hemir-ings, and pre-semirings are commutative, unless otherwise is stated.

1. Finitely Additive Functions on Hemirings

Let us recall that if X is a fixed set, then an algebra of sets A is a class of subsetsofX such that ∅, X ∈ A and if A,B ∈ A, then A∪B ∈ A, A∩B ∈ A, and A′ ∈ A. Inmeasure theory, a real-valued function µ on an algebra A is called a finitely additivefunction if for disjoint subsets A and B of X , we have µ(A ∪ B) = µ(A) + µ(B)[4]. From this definition, we are inspired to define finitely additive functions overhemirings as follows:

Definition 1.1. Let S be a hemiring and s, t, s1, s2, . . . , sn ∈ S.

(1) The two elements s and t are said to be disjoint, if st = 0.(2) The n elements s1, s2, . . . , sn (n > 1) are said to be mutually disjoint, if

sisj = 0 for all 1 ≤ i < j ≤ n.(3) Let T be a commutative semigroup. We define a function f : S −→ T to

be finitely additive if for any disjoint elements s and t of S, we have

f(s+ t) = f(s) + f(t).

Let us mention that there are plenty of examples for finitely additive functionson various algebraic structures defined in different branches of mathematics. Sincea modular function m defined over a hemiring with m(0) = 0 is also a finitelyadditive function (refer to Theorem 3.24), those examples given in Example 3.21 arefinitely additive functions as well. Probability functions are also finitely additive.Therefore, the functions, given in Example 2.2, are also interesting finitely additivefunctions. By the way, we give a couple of interesting examples for finitely additivefunctions over hemirings in the following:

Example 1.2. Examples of finitely additive functions over hemirings:

(1) Let I be the collection of finite unions of all intervals of the form [c, d],[c, d), (c, d], and (c, d) where c, d ∈ R. It is easy to see that (I,∪,∩, ∅) isa hemiring. For any interval J of the form [c, d], [c, d), (c, d], and (c, d),set µ(J) = |d− c| and for a finite union of disjoint intervals J1, J2, . . . , Jn,set µ(

⋃ni=1 Ji) =

∑ni=1 µ(Ji). Then one can easily check that µ is a finitely

additive function on the hemiring I [4, Chap. 1].(2) Let X be an infinite set and A be the algebra of sets B ⊆ X such that either

B or Bc is finite (in some references, when Bc is finite, B is called co-finite).


For finite B, define f(B) = 0, and for co-finite B, define f(B) = 1. It iseasy to see that (A,∪,∩, ∅, X) is a Boolean algebra and therefore a hemiringand f a finitely additive function [3, Problem 2.13].

(3) Let T be a commutative semigroup and E an entire hemiring, that is ab = 0implies that either a = 0 or b = 0 for any a, b ∈ E. Any function f : E −→ Twith the property that f(0) = 0 is finitely additive and the reason is asfollows: If a, b ∈ E and ab = 0, then either a = 0 or b = 0 and in each case,f(a+ b) = f(a) + f(b).

Proposition 1.3. Let S be a hemiring, T a commutative semigroup, and f : S −→T finitely additive. Then for any disjoint elements s1, s2, . . . , sn ∈ S, we have

f(s1 + s2 + · · ·+ sn) = f(s1) + f(s2) + · · ·+ f(sn).

Proof. By Definition 1.1, the statement holds for n = 2. Now let n > 2 and supposethat s1, . . . , sn are elements of S such that sisj = 0 for all 1 ≤ i < j ≤ n. It is clearthat

(s1 + · · ·+ sn−1) · sn = 0

and therefore

f((s1 + · · ·+ sn−1) + sn) = f(s1 + · · ·+ sn−1) + f(sn).

But by induction’s hypothesis,

f(s1 + · · ·+ sn−1) = f(s1) + · · ·+ f(sn−1),

and the proof is complete.

Example 1.2 shows that a finitely additive function f : S −→ T , where S is ahemiring, will be more interesting if S is not an entire hemiring, i.e., a hemiringwith nontrivial zero-divisors. Semirings with nontrivial complemented elements areinteresting examples of semirings with nontrivial zero-divisors and suitable for ourpurpose in this paper. Let us recall that an element s of a semiring S is said to becomplemented if there exists an element cs ∈ S satisfying s · cs = 0 and s+ cs = 1.The mentioned element cs ∈ S is called the complement of s. One can easily checkthat if s ∈ S has a complement, then it is unique. The complement of s ∈ S, if itexists, is denoted by s⊥. Also, note that the set of all complemented elements ofa semiring S is denoted by comp(S). The set comp(S) is always nonempty, since0 ∈ comp(S). It is clear that if s ∈ comp(S), then s⊥ ∈ comp(S) and (s⊥)⊥ = s.Finally if s, t ∈ S are complemented, the symmetric difference of s and t is definedto be st = s⊥t+ st⊥ [12, Chap. 5].

Proposition 1.4. Let S be a semiring, T a commutative semigroup, and f : S −→T a finitely additive function. If s ∈ S is complemented, then the following state-ments hold:

(1) f(t) = f(ts) + f(ts⊥), for any t ∈ S.(2) If, in addition, t ∈ S is also complemented, we have

f(st) + 2f(st) = f(s) + f(t).

Proof. (1): Since (ts)(ts⊥) = 0, we have

f(t) = f(t(s+ s⊥)) = f(ts+ ts⊥) = f(ts) + f(ts⊥).


(2): Since (s⊥t)(st⊥) = 0, we have the following:

f(st)+2f(st) = f(s⊥t+st⊥)+2f(st) = f(s⊥t)+f(st)+f(st⊥)+f(st) = f(t)+f(s),


A normalized measure in measure theory is a measure µ with µ(1) = 1 [17, p.65]. Now it is natural to give the following definition. In current and the nextsection, we are especially interested in finitely additive normalized functions.

Definition 1.5. Let S and R be semirings. We define a function f : S −→ R tobe normalized if f(1S) = 1R.

Proposition 1.6. Let S and T be semirings and f : S −→ T a finitely additivefunction such that f(0) = 0. Then the following statements are equivalent:

(1) The function f is normalized.(2) For any complemented element s ∈ S, f(s) + f(s⊥) = 1.

Proof. (1) ⇒ (2): Since s ∈ S is complemented, ss⊥ = 0. Obviously this impliesthat f(s) + f(s⊥) = f(s+ s⊥) = f(1) = 1.

(2) ⇒ (1): Since 0 ∈ comp(S), by assumption f(0) + f(0⊥) = 1. But f(0) = 0and 0⊥ = 1. Therefore, f(1) = 1.

Note that in Proposition 1.6, it is possible to assume that T is a ring insteadof assuming that f(0) = 0, since cancellation law of addition in rings implies thatf(0) = 0 automatically (Refer to Remark 3.2). Let us recall that the operation ⊔is defined as s ⊔ t = s+ s⊥t, where s ∈ comp(S) and t ∈ S [12, Chap. 5].

Proposition 1.7. Let S be a zerosumfree semiring, R a ring, and f : S −→ R afinitely additive and normalized function. If s and t are complemented elements ofS, then

f(s⊥t⊥) = 1− f(s)− f(t) + f(st).

Proof. Let S be a zerosumfree semiring and s, t ∈ comp(S). According to theproof of Proposition 5.6 in [12], we have s⊥t⊥ = (s ⊔ t)⊥ and s ⊔ t ∈ comp(S).So by Proposition 1.4, we have: f(s⊥t⊥) = 1 − f(s ⊔ t) = 1 − f(s + s⊥t) =1− f(s)− f(s⊥t) = 1− f(s)− f(t) + f(st).

Now we give the following definition inspired from the concept of independentevents in probability theory [9, p. 115].

Definition 1.8. Let S and R be pre-semirings and f : S −→ R a function. Wedefine s1, s2, . . . , sn ∈ S to be independent, if

f(∏

x∈X

x) =∏

x∈X

f(x),

for any nonempty subset X of s1, s2, . . . , sn.

Theorem 1.9. Let S be a semiring, R a ring, and f : S −→ R a finitely additiveand normalized function. Then the following statements for complemented elementss1, s2, . . . , sn of S are equivalent:

(1) The elements s1, s2, . . . , sn are independent,(2) The 2n sets of elements of the form t1, t2, . . . , tn with ti = si or ti = s⊥i are

independent,


(3) The 2n equalities

f(t1t2 · · · tn) = f(t1)f(t2) · · · f(tn)

with ti = si or ti = s⊥i hold.

Proof. (1) ⇒ (2): Let t1, . . . , tm ⊆ s2, . . . , sn. By Proposition 1.4, we have

f(t1 · · · tm) = f(t1 · · · tms1) + f(t1 · · · tms⊥1 ).

But s1, t1, . . . , tm are independent. So we have

f(t1) · · · f(tm) = f(s1)f(t1) · · · f(tm) + f(t1 · · · tms⊥1 ).

This implies that

f(t1 · · · tms⊥1 ) = f(t1) · · · f(tm)− f(s1)f(t1) · · · f(tm) =

(1− f(s1))f(t1) · · · f(tm) = f(s⊥1 )f(t1) · · · f(tm).

This means that s⊥1 and s2, . . . , sn are independent. Now it is clear that bymathematical induction, we can prove that s⊥1 , . . . , s

⊥i , si+1, . . . , sn are also inde-

pendent.(2) ⇒ (3): Obvious.(3) ⇒ (1): Note that since f(s) + f(s⊥) = 1 for any complemented element of

s ∈ S,

f(s1)f(s2) · · · f(sm) = f(s1)f(s2) · · · f(sm)(f(sm+1)+f(s⊥m+1)) · · · (f(sn)+f(s⊥n )).

From the 2n equalities f(t1 · · · tn) = f(t1) · · · f(tn) with ti = si or ti = s⊥i , we getthat

f(s1)f(s2) · · · f(sm) =

f(s1 · · · smsm+1 · · · sn) + f(s1 · · · sms⊥m+1 · · · sn) + · · ·+ f(s1 · · · sms⊥m+1 · · · s⊥n ).

But the elements s1 · · · smsm+1 · · · sn, s1 · · · sms⊥m+1 · · · sn, . . . and s1 · · · sms⊥m+1 · · · s⊥n

are mutually disjoint, so by using Proposition 1.3, we have:

f(s1)f(s2) · · · f(sm) =

f(s1 · · · smsm+1 · · · sn + s1 · · · sms⊥m+1 · · · sn + · · ·+ s1 · · · sms⊥m+1 · · · s⊥n ) =

f(s1 · · · sm(sm+1 + s⊥m+1) · · · (sn + s⊥n )) = f(s1 · · · sm).

The thing we wished to show.

Let us recall that a function d : X ×X −→ R+ ∪ +∞ is defined to be semi-metric, if it satisfies all properties of a metric function except the requirement thatd(x, y) = 0 implies x = y. In other words, it is allowed that distinct points may havea zero distance [5, Definition 1.4.4.]. We give a similar definition for G-semi-metricfunctions, where G is an ordered Abelian group.

Definition 1.10. Let X be an arbitrary set and (G,+,≤) be an ordered Abeliangroup.

(1) A function f : X −→ G is defined to be non-negative if f(x) ≥ 0 for anyx ∈ X . If (X,+, 0) is a monoid, then the function f is defined to be positiveif f is non-negative and f(x) = 0 if and only if x = 0, for any x ∈ X .

(2) A function d : X × X −→ G defines a G-semi-metric on X , if d has thefollowing properties:(a) Non-negativeness: d(x, y) ≥ 0 and d(x, x) = 0, for any x, y ∈ X .(b) Symmetry: d(x, y) = d(y, x), for any x, y ∈ X .


(c) Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z), for any x, y, z ∈ X .Also, d : X × X −→ G defines a G-metric on X , if d defines a G-

semi-metric on X and d(x, y) = 0 implies that x = y, for any x, y ∈ X(Positiveness).

Theorem 1.11. Let S be a zerosumfree semiring, G an ordered Abelian group,and f : S −→ G a non-negative and finitely additive function. Then the functiond(s, t) = f(st) defines a G-semi-metric on comp(S). Moreover, if the functionf : S −→ G is positive, then d defines a G-metric on comp(S).

Proof. Let s, t, u ∈ comp(S). It is clear that d(s, t) ≥ 0 and d(s, s) = f(ss) =f(0) = 0. Also, the symmetric property can be shown easily: d(s, t) = f(st) =f(ts) = d(t, s). Now we show the triangle inequality holds, that is, d(s, t) +d(t, u) ≥ d(s, u).

d(s, t) + d(t, u)− d(s, u) = f(st) + f(tu)− f(su)= f(s⊥t) + f(st⊥) + f(t⊥u) + f(tu⊥)− f(s⊥u)− f(su⊥)= f(s⊥tu) + f(s⊥tu⊥) + f(st⊥u) + f(st⊥u⊥) + f(st⊥u) + f(s⊥t⊥u)+f(stu⊥) + f(s⊥tu⊥)− f(s⊥tu)− f(s⊥t⊥u)− f(stu⊥)− f(st⊥u⊥)= 2[f(st⊥u) + f(s⊥tu⊥)] ≥ 0.Finally, let the function f : S −→ G be positive and assume that d(s, t) = 0.

Then f(s⊥t) + f(st⊥) = 0. From this we get that f(s⊥t) = f(st⊥) = 0, which inturn implies that s⊥t = st⊥ = 0.

Now s = s · 1 = s(t⊥ + t) = st⊥ + st = st. In a similar way, t = ts and therefores = t and this completes the proof.

2. Probability Functions on Semirings

Let us recall that if A is an algebra of sets over a fixed set X , by a probabilityfunction, it is meant a non-negative, normalized, and finitely additive function [21,Chap. I]. Inspired by this, we define probability functions over semirings similarly.Note that by an ordered semiring, we mean a semiring (S,+, ·) with a partial order≤ on S such that the following statements hold:

(1) If s ≤ t, then s+ u ≤ t+ u for any s, t, u ∈ S;(2) If s ≤ t and 0 ≤ u, then su ≤ tu for any s, t, u ∈ S.

For more on ordered semirings, one may refer to [11, Chap. 2].

Definition 2.1. Let S be a semiring and T an ordered semiring. We define afunction p : S −→ T to be a probability function, if the following properties aresatisfied:

(1) p(s) ≥ 0, for any s ∈ S.(2) p(1) = 1.(3) If s, t ∈ S and st = 0, then p(s+ t) = p(s) + p(t).

Example 2.2. Here are a couple of examples for probability functions:

(1) Let X be a nonempty finite or countable set. It is clear that the Booleanalgebra (P (X),∪,∩) has a semiring structure. Let pxx∈X be a stochasticsequence, that is, px are non-negative real numbers with

∑

x px = 1. Onemay define a function p : P (X) −→ R with p(A) =

∑

k∈A pk. It is a well-known fact in probability theory that p defines a probability function onthe semiring P (X) [16].


(2) Let a and b be two real numbers such that a < b and Iab = [a, b] the closedinterval of real numbers. A set Y ⊆ Iab is called to be a subinterval ofIab if Y is in one of the forms [c, d], [c, d), (c, d], or (c, d), where a ≤ c ≤d ≤ b. One may define S to be the collection of all possible finite unionsof subintervals of Iab. It is, then, easy to check that (S,∪,∩) is a semiring.Now we define p : S −→ R in the following way:

(a) p(I) =d− c

b− a, if I is one of the subintervals [c, d], [c, d), (c, d], or (c, d)

of Iab, where a ≤ c ≤ d ≤ b.(b) p(J) = p(I1) + · · ·+ p(Im), where J = I1 ∪ · · · ∪ Im and I1, . . . , Im are

distinct subintervals of Iab such that Iα ∩ Iβ is either the empty set ora singleton for any 1 ≤ α < β ≤ m.

It is, then, easy to see that p defines a probability function on the semir-ing (S,∪,∩) [9].

Proposition 2.3. Let S be a semiring, T an ordered ring, and p : S −→ T aprobability function. If s ∈ S is complemented, then the following statements hold:

(1) 0 ≤ p(s) ≤ 1.(2) p(ts) ≤ p(t), for any t ∈ S.

Moreover, if S is a zerosumfree semiring and s, t ∈ S are both complemented,then we have the following:

(3) p(st) ≥ p(s) + p(t)− 1.

Proof.(1): 0 ≤ p(s) ≤ p(s) + p(s⊥) = 1.(2): Since p(t) = p(ts) + p(ts⊥), we have that p(t) − p(ts) = p(ts⊥) ≥ 0, which

means that p(ts) ≤ p(t).(3): By Proposition 1.7, we have p(s⊥t⊥) = 1 − p(s) − p(t) + p(st). Since

p(s⊥t⊥) ≥ 0, p(st) ≥ p(s) + p(t)− 1.

Let us recall that if s ∈ S is complemented, then ⊔ is defined as s⊔ t = s+ s⊥t.Now we prove the following theorem related to complemented elements of a semiring[12, Chap. 5].

Theorem 2.4. Let S be a zerosumfree semiring. Then the following statementsare equivalent:

(1) If s, t ∈ comp(S), then s+ t ∈ comp(S),(2) 1 + 1 ∈ comp(S),(3) If s, t ∈ comp(S), then s+ t = s ⊔ t, for any s, t ∈ S,(4) (comp(S),+, ·) is a Boolean algebra.

Proof. The assertion (1) ⇔ (2) ⇔ (3) holds by [12, Proposition 5.7].(3) ⇒ (4): By Corollary 5.9 in [12], if S is a zerosumfree semiring, then (comp(S),⊔, ·)

is a Boolean algebra. But by assumption s+t = s⊔t, for any s, t ∈ comp(S). There-fore (comp(S),+, ·) is a Boolean algebra.

(4) ⇒ (3): Straightforward.

Theorem 2.5 (Equality of Parallel Systems). Let S be a zerosumfree semiring suchthat 1 + 1 ∈ comp(S), T a ring, and f : S −→ T a finitely additive and normalized


function. If s1, s2, . . . , sn ∈ S are complemented and independent, then

f

(

n∑

i=1

si

)

= 1−n∏

i=1

(1− f(si)).

Proof. Since S is a zerosumfree semiring such that 1 + 1 ∈ comp(S), by Theorem2.4, (comp(S),+, ·) is a Boolean algebra. But s1, s2, . . . , sn ∈ S are complemented,so it is clear that the complement of

∑ni=1 si is

∏ni=1 s

⊥i and we have

n∑

i=1

si +

n∏

i=1

s⊥i = 1.

Since f is a finitely additive and normalized function, we have

f

(

n∑

i=1

si

)

+ f

(

n∏

i=1

s⊥i

)

= 1.

On the other hand, s1, s2, . . . , sn ∈ S are independent, so by Theorem 1.9,

f

(

n∏

i=1

s⊥i

)

=

n∏

i=1

f(s⊥i ).

But f(s⊥i ) = 1− f(si) and this finishes the proof.

An important technique in set theory, known colloquially as “disjointification,”says that if A1, A2, . . . , An are arbitrary sets, then the sets defined by Bk = Ak \⋃k−1

i=1 Ai are mutually disjoint and⋃n

i=1 Bi =⋃n

i=1 Ai [20, Proposition 1.24]. Thistechnique can be generalized in any Boolean algebra:

Lemma 2.6 (Disjointification Technique). Let (B,+, ·, 0, 1,′ ) be a Boolean algebraand a1, a2, . . . , an ∈ B. Then the elements bk = aka

′1 · · · a

′k−1 are mutually disjoint

and b1 + · · ·+ bn = a1 + · · ·+ an.

Proof. It is straightforward to see that bis are mutually disjoint. Now we proceedby induction. It is clear that the assertion holds for n = 1. Let the assertionhold for n = k and we prove it for the case n = k + 1. Let a1, a2, . . . , ak+1 ∈ Band set bl = ala

′1 · · ·a

′l−1 for any 1 ≤ l ≤ k + 1. By induction’s hypothesis,

b1 + · · ·+ bk = a1 + · · ·+ ak, so

b1 + b2 + · · ·+ bk + bk+1 = a1 + a2 + · · ·+ ak + ak+1a′1 · · ·a

′k.

Now if we set s = a1 + a2 + · · ·+ ak, we have that s′ = a′1a′2 · · · a

′k, and finally

b1 + b2 + · · ·+ bk + bk+1 = s+ ak+1s′ = (s+ ak+1)(s+ s′) = s+ ak+1.

So by induction, the proof is complete.

Theorem 2.7 (Boole’s Inequality). Let S be a zerosumfree semiring such that1 + 1 ∈ comp(S), T an ordered ring, and p : S −→ T a probability function. Ifs1, s2, . . . , sn ∈ S are complemented, then

p(s1 + s2 + · · ·+ sn) ≤ p(s1) + p(s2) + · · ·+ p(sn).


Proof. By Theorem 2.4, (comp(S),+, ·) is a Boolean algebra. Let s1, s2, . . . , sn ∈comp(S) and define t1 = s1 and tk = sks

⊥1 · · · s⊥k−1 for any 1 < k ≤ n. It is clear

that t1, t2, . . . , tn are mutually disjoint elements of comp(S) and by Lemma 2.6,t1 + t2 + · · ·+ tn = s1 + s2 + · · ·+ sn. Moreover, by Proposition 2.3, p(tk) ≤ p(sk).Therefore,

p(s1 + · · ·+ sn) = p(t1 + · · ·+ tn) = p(t1) + · · ·+ p(tn) ≤ p(s1) + · · ·+ p(sn),


Definition 2.8. Let S be a semiring, T an ordered semiring, and p : S −→ T aprobability function. If for t ∈ S, p(t) is a multiplicatively invertible element of thesemiring T , the “conditional probability of s given t,” denoted by p(s|t), is definedto be p(s|t) = p(st)/p(t).

The proof of the following proposition is straightforward.

Proposition 2.9. Let S be a semiring, T an ordered semiring, and p : S −→ T aprobability function. If for the elements s, t ∈ S, p(s) and p(t) are multiplicativelyinvertible elements of the semiring T , then the following statements are equivalent:

(1) p(s|t) = p(s),(2) p(t|s) = p(t),(3) The elements s and t are independent, i.e., p(st) = p(s)p(t).

Proposition 2.10. Let S be a semiring, T an ordered semiring, and p : S −→ Ta probability function. If for an element t ∈ S, p(t) is a multiplicatively invertibleelement of the semiring T , then pt : S −→ T defined by pt(s) = p(s|t) is a probabilityfunction, i.e., the following statements hold:

(1) pt(s) ≥ 0 for any s ∈ S.(2) pt(1) = 1.(3) If s1s2 = 0, for s1, s2 ∈ S then pt(s1 + s2) = pt(s1) + pt(s2).

Proof. Straightforward.

Proposition 2.11 (Law of Total Probability). Let S be a semiring, T an orderedsemiring, and p : S −→ T a probability function. If t1, t2, . . . , tn are elements ofS such that titj = 0 for any 1 ≤ i < j ≤ n, t1 + t2 + · · · + tn = 1, and p(ti) is amultiplicatively invertible element of the semiring T for any 1 ≤ i ≤ n, then

p(s) = p(s|t1)p(t1) + p(s|t2)p(t2) + · · ·+ p(s|tn)p(tn).

Proof. It is clear that s = st1 + st2 + · · · + stn and (sti)(stj) = 0 for any 1 ≤ i <j ≤ n. So by Proposition 1.3, we have that p(s) = p(st1)+p(st2) · · ·+p(stn). Sincep(sti) = p(s|ti)p(ti) for any 1 ≤ i ≤ n,

p(s) = p(s|t1)p(t1) + · · ·+ p(s|tn)p(tn),


Corollary 2.12 (Law of Total Probability). Let S be a semiring, T an orderedsemiring, and p : S −→ T a probability function. If t ∈ S is a complementedelement of S such that p(t) and p(t⊥) are multiplicatively invertible elements of thesemiring T , then p(s) = p(s|t)p(t) + p(s|t⊥)p(t⊥).


Proposition 2.13. Let S be a semiring, T an ordered semiring, and p : S −→ Ta probability function. If for the elements s ∈ S, t ∈ comp(S), p(s), p(t), andp(t⊥) are multiplicatively invertible elements of the semiring T , then the followingstatements are equivalent:

(1) p(s|t) = p(s|t⊥),(2) The elements s and t are independent.

Proof. (1) ⇒ (2): By the Law of Total Probability, we have

p(s) = p(s|t)p(t) + p(s|t⊥)p(t⊥) =

p(s|t)p(t) + p(s|t)p(t⊥) = p(s|t)(p(t) + p(t⊥)) = p(s|t).

(2) ⇒ (1): Since the elements s and t are independent, by Proposition 2.9, wehave that p(s) = p(s|t) and p(s) = p(s|t⊥).

A purpose of Bayes’ formula in probability theory is to compute P (B|A) interms of P (A|B). We prove a semiring version of Bayes’ Theorem for probabilityfunctions as follows:

Theorem 2.14 (Bayes’ Theorem). Let S be a semiring, K an ordered semifield,and p : S −→ K a probability function. If t1, . . . , tn are elements of S such thattitj = 0 for any 1 ≤ i < j ≤ n, t1 + · · ·+ tn = 1, p(ti) > 0 for any 1 ≤ i ≤ n, ands ∈ S such that p(s) > 0, then

p(tk|s) =p(s|tk)p(tk)

p(s|t1)p(t1) + · · ·+ p(s|tn)p(tn).

Proof. By definition, we know that p(tk|s) =p(tks)

p(s). Also, p(tks) = p(stk) =

p(s|tk)p(tk). On the other hand, by the Law of Total Probability (Proposition2.11), we have

p(s) = p(s|t1)p(t1) + · · ·+ p(s|tn)p(tn).

Therefore,

p(tk|s) =p(s|tk)p(tk)

p(s|t1)p(t1) + · · ·+ p(s|tn)p(tn),


3. Modular Functions on Pre-semirings

Modular functions appear in different branches of mathematics. A modularfunction is usually a real-valued function m over some objects that two operations“ + ” and “ · ” have meaning for those objects and if s and t are of those objects,we have the following equality:

m(s+ t) +m(st) = m(s) +m(t).

In this section, we define modular functions on pre-semirings and examine modularfunctions over different pre-semirings mentioned in literature and show that mod-ular functions over these pre-semirings are almost constant, i.e., they are constantover their domains except for a finite number of points in them. We also prove thatthere are many modular functions over the semiring (Id(D),+, ·), where D is aDedekind domain. After that, we give a couple of nontrivial examples for modular


functions in Remark 3.2 and Example 3.21. Then we prove Poincare’s Inclusion-Exclusion Theorem for modular functions over pre-semirings. We also prove thatover complemented elements of a semiring, finitely additive functions are modular.

Definition 3.1. Let S be a pre-semiring and T a commutative semigroup. Wedefine a function m : S −→ T to be modular if m(s+ t) +m(st) = m(s) +m(t) forall s, t ∈ S.

Remark 3.2.

(1) If S is a pre-semiring such that 0 and 1 are the additive and multiplicativeneutral elements of S, respectively, T is a commutative and cancellativemonoid, and f : S −→ T is a modular function, then by choosing t = 1,from the following modular equality

f(s+ t) + f(st) = f(s) + f(t),

we get that f(s + 1) = f(1). This means that f is constant on the basicsub-pre-semiring 1, 1+1, 1+1+1,. . . of S.

(2) Let S be a hemiring, T a commutative monoid, and m : S −→ T a function.It is clear that if m is modular and m(0) = 0, then m is finitely additive.The question arises when the inverse is correct. Note that if T is a can-cellative monoid and f is finitely additive then f(0) = 0 and the proof is asfollows: since 0 · 0 = 0, we have that f(0 + 0) = f(0) + f(0) and since T iscancellative, f(0) = 0. In Theorem 3.24, we show that in Boolean algebrasmodularity of a function f is caused by its finitely additive property.

(3) Let S be a totally ordered set. It is easy to check that (S,⊕,⊙) is apre-semiring (known as bottleneck pre-semiring), where a⊕ b = maxa, band a ⊙ b = mina, b for all a, b ∈ S. Now let (T,+) be a commutativesemigroup. Obviously for any function m : S −→ T , we have

m(x⊕ y) +m(x⊙ y) = m(x) +m(y).

Now we start examining modular functions over different pre-semirings. Let N0

denote the set 0, 1, 2, . . . , k, k+1, . . ., i.e., the set of all nonnegative integers. Onemay define addition and multiplication over S = N0 ∪ −∞ as “max” and “ + ”respectively by considering that −∞ < n < n+1 for all n ∈ N0 and −∞+ s = −∞for all s ∈ S. It is easily checked that (N0 ∪ −∞,max,+,−∞, 0) is a semiringand in some references, it is known as the Arctic semiring [8, p. 179].

Proposition 3.3. Let f : N0∪−∞ −→ G be a function from the Arctic semiring(N0∪−∞,max,+,−∞, 0) to the Abelian group G. Then f is a modular functionif and only if it is a constant function over N0.

Proof. ⇒: By modularity of the function f , we have:

f(maxx, y) + f(x+ y) = f(x) + f(y).

If we suppose that x ≥ 0 and y = 0, we get that f(x) + f(x) = f(x) + f(0) andthis means that f is constant over N0.

⇐: On the other hand, if f is constant over N0, then f is modular over theArctic semiring.

Let us recall that the semiring (Tk,max,mina+ b, k,−∞, 0), where 1 ≤ k andTk = −∞, 0, 1, . . . , k is called the Truncation semiring .


Proposition 3.4. Let G be an Abelian group. Then f : Tk −→ G is modular ifand only if f is constant over Tk − −∞, where by Tk, we mean the Truncationsemiring.

Proof. ⇒: Let Tk be the Truncation semiring, G an Abelian group, and f : Tk −→G a modular function. So by definition, we have

f(maxx, y) + f(minx+ y, k) = f(x) + f(y),

for all x, y ∈ Tk and if we let y = k, we have the following:

f(k) + f(minx+ k, k) = f(x) + f(k).

This implies that f(minx+ k, k) = f(x). But minx+ k, k = k, if x ≥ 0. So fis constant over Tk − −∞.

⇐: Straightforward.

Another important family of finite semirings is mentioned in [12, Example 1.8]and we bring it here for the convenience of the reader. Let us recall that if i and nare positive integers such that i < n, the addition and multiplication of the finitesemiring

(

B(n, i) = 0, 1, . . . , n− 1,⊕,⊙, 0, 1)

are defined as follows:The addition ⊕ is defined as x ⊕ y = x + y if x + y ≤ n − 1 and x ⊕ y = l if

x+ y > n− 1 where l is the unique number satisfying the conditions i ≤ l ≤ n− 1and l ≡ x+ y (mod n− i) and multiplication ⊙ is defined similarly. Now we provethe following interesting result:

Proposition 3.5. Let G be an Abelian group and i < n be positive integers. Then afunction f : B(n, i) −→ G is modular if and only if it is constant over B(n, i)−0.

Proof. ⇒: Let y = 1. It is clear that if x < n− 2, then x+ y and xy are both lessthan n− 1. So x⊕ 1 = x+ 1, and x⊙ 1 = x1 = x.

Putting y = 1 and x < n− 2 in the modular equation, we have

f(x+ 1) + f(x) = f(x) + f(1),

which impliesf(x+ 1) = f(1),

for x = 0, 1, . . . , n− 2. Thus f(x) = f(1) for x = 1, 2, . . . , n− 1. This means thatf is constant over B(n, i)− 0.

⇐: Obvious.

Proposition 3.6. Let (G,+,≤) be a totally ordered Abelian group and T a com-mutative and cancellative monoid. Then (G,min,+) is a pre-semirng and the onlymodular function, m : G −→ T , over the pre-semiring G is a constant function.

Proof. It is straightforward to check that (G,min,+) is a pre-semiring. Note thatby assumption, the following equality holds for all x, y ∈ G:

m(minx, y) +m(x + y) = m(x) +m(y).

Now take x ≥ 0. It is, then, clear that x ≥ −x and minx,−x = −x. So bymodularity of m over the pre-semiring G, we have m(−x) +m(0) = m(x) +m(−x)and therefore m(x) = m(0). On the other hand, if we let x < 0, then minx, 0 = xand it is clear that the modularity ofm implies thatm(x)+m(x) = m(x)+m(0) and


again we have m(x) = m(0). This means that for any x ∈ G, we have m(x) = m(0)and m is a constant function and the proof is complete.

Proposition 3.7. Let S be a pre-semiring such that 0 and 1 are the additive andmultiplicative neutral elements of S, respectively. If 1 has an additive inverse, Tis a commutative and cancellative monoid, and a function f : S −→ T is modular,then f is a constant function.

Proof. Let x ∈ S be arbitrary and set y = 1. So by modularity, we have:

f(x+ 1) + f(x · 1) = f(x) + f(1).

Since T is a cancellative monoid, we have f(x+1) = f(1). Now since 1 has additiveinverse, we can replace x by x− 1 and we have f(x) = f(1), which means that thefunction f is constant on S and this finishes the proof.

Corollary 3.8. If R is a commutative ring with identity, T a commutative andcancellative monoid, and a function f : S −→ T is modular, then f is a constantfunction.

Remark 3.9. Note that if the multiplicative neutral element 1 of a semiring R hasan additive inverse, i.e., 1+(−1) = 0, then by distributive law, we have r+(−1)r = 0for any r ∈ R and this means that R is a ring. Therefore, in order to show thatProposition 3.7 is really a generalization of Corollary 3.8, it is important to give anexample of a pre-semiring S such that 0 and 1 are the additive and multiplicativeneutral elements of S, respectively, and 1 has an additive inverse, while S is not asemiring. In fact, the interesting example given in [14, Example 5.3.1] fulfills ourpurpose and inspires us to give the following result:

Proposition 3.10. Let (H,+, 0,−,≤) be a nontrivial totally ordered Abelian groupand E be the set of all closed intervals [a, b] ⊆ H such that a ≤ 0 and b ≥ 0. Define⊕ and ⊗ on E by

[a1, b1]⊕ [a2, b2] = [mina1, a2,maxb1, b2],

[a1, b1]⊗ [a2, b2] = [a1 + a2, a2 + b2],

for all a1, a2 ≤ 0 and b1, b2 ≥ 0. Then the following statements hold:

(1) (E,⊕,⊗) is a pre-semiring and [0, 0] is the neutral element for both opera-tions ⊕ and ⊗.

(2) The neutral element for multiplication ⊗ in E has an additive inverse.(3) The neutral element [0, 0] of addition ⊕ is not an absorbing element of E

and so E is not a semiring.(4) If G is an Abelian group, then f : E −→ G is modular if and only if f is

constant.

Proof.(1) and (2): Straightforward.(3): [a, b]⊗ [0, 0] = [a, b] for all [a, b] ∈ E.(4): This holds by Proposition 3.7.

Proposition 3.11. Let G be an Abelian group and R = (R× R) ∪ −∞. Defineoperations of R as follows:

(1) (a, b)⊕ (a′, b′) = (maxa, a′,maxb, b′) for all a, b, a′, b′ ∈ R;(2) (a, b)⊙ (a′, b′) = (a+ a′, b+ b′) for all a, b, a′, b′ ∈ R;


(3) (−∞)⊕ r = r ⊕ (−∞) = r for all r ∈ R;(4) (−∞)⊙ r = r ⊙ (−∞) = −∞ for all r ∈ R.

Then (R,⊕,⊙) is a semiring and any modular function f : R −→ G is constant.

Proof. According to [12, Example 1.25], R is a semiring and it is clear that 1 = (0, 0)is the identity element of the multiplication of R. Let x, y ∈ R. By modularity off , we have

f((x, y)⊕ (0, 0)) + f((x, y)⊙ (0, 0)) = f(x, y) + f(0, 0),

which means that

f(maxx, 0,maxy, 0) = f(0, 0),

and clearly this implies that if x, y ≥ 0, then

f(x, y) = (0, 0).

On the other hand, since (x, y) ⊕ (−x,−y) = (|x|, |y|) and (x, y) ⊙ (−x,−y) =(0, 0), modularity of f implies that

f(|x|, |y|) + f(0, 0) = f(x, y) + f(−x,−y).

Therefore, if we let x, y ≥ 0, we have

f(−x,−y) = f(0, 0).

Now, let x, y ≥ 0. Then it is clear that (x, y) ⊕ (x,−y) = (x, y) and (x, y) ⊙(x,−y) = (2x, 0) and again by modularity of f , we have

f(x, y) + f(2x, 0) = f(x, y) + f(x,−y).

Also, f(2x, 0) = f(0, 0). So, f(x,−y) = f(0, 0). Similarly, one can see thatf(−x, y) = f(0, 0) and the proof is complete.

Proposition 3.12. Let (G,+) be an Abelian group. Then the following statementsfor a function f : N0 ∪ +∞ −→ G are equivalent:

(1) The function f : N0 ∪ +∞ −→ G is a modular function from the tropicalalgebra (N0 ∪ +∞,min,+) to the Abelian group G,

(2) The function f is constant over N.

Proof. (1) ⇒ (2): Let y = 1 and x ∈ N and rewrite the modular equality for thetropical algebra:

f(minx, 1) + f(x+ 1) = f(x) + f(1).

Since x ∈ N, minx, 1 = 1, we have f(minx, 1) = f(1), which gives

f(1) + f(x+ 1) = f(x) + f(1).

Canceling f(1) from both sides, we have f(x+1) = f(x) for all x ∈ N, which meansf is constant over N.

(2) ⇒ (1): Since f is constant over N, we may choose constants k, c, l ∈ G suchthat

f(x) =

k if x = 0

c if x ∈ N

l if x = +∞

It is straightforward to see that the function f satisfies the modular equality andthis finishes the proof.


Theorem 3.13. Let G be an Abelian group and (K,+, ·,≤) a totally ordered posi-tive semifield with the property that for y ≥ 1, there is an x ≥ 0 such that y = x+1for all x, y ∈ K. Then the following statements for the function f : K −→ G areequivalent:

(1) The function f is modular,(2) The function f is constant over the set of positive elements X = x > 0 :

x ∈ K of K.

Proof. (2) ⇒ (1): Straightforward.(1) ⇒ (2): Since f is a modular function, it satisfies the following equation:

f(x+ y) + f(xy) = f(x) + f(y).

Let y = 1 and rewrite the functional equation:

f(x+ 1) + f(x) = f(x) + f(1),

which implies that f(x + 1) = f(1) for all x ≥ 0. Since for all y ≥ 1, there is anx ≥ 0 such that y = x+ 1, we have that

(3.1) f(x) = f(1) ∀x ≥ 1.

Now replace y =1

xin the original equation to obtain

(3.2) f

(

x+1

x

)

+ f(1) = f(x) + f

(

1

x

)

.

Assume 0 < x ≤ 1, then1

x≥ 1 and from equation (3.1), we have f

(

1

x

)

= f(1).

Replacing this in equation (3.2), we get f

(

x+1

x

)

+f(1) = f(x)+f(1). Removing

f(1) from both sides, we have

f

(

x+1

x

)

= f(x) ∀ 0 < x ≤ 1.

Since x +1

x≥ 1, we have f

(

x+1

x

)

= f(1). Therefore, f(x) = f(1) for all

0 < x ≤ 1. Finally, f(x) = f(1) = c for some constant c ∈ T and all x ∈ X and theproof is complete.

Example 3.14. In order to show the importance of Theorem 3.13, let us give acouple of nice examples for semifields satisfying the conditions of Theorem 3.13.

(1) It is clear that (Q≥0,+, ·,≤) is a totally ordered positive semifield with theproperty that for y ≥ 1, there is an x ≥ 0 such that y = x + 1 for allx, y ∈ Q≥0.

(2) It is easy to check that if (K,+, ·,≤) is a totally ordered positive semifield,then the semifield (K,max, ·,≤) is a totally ordered positive semifield withthe property that for y ≥ 1, there is an x ≥ 0 such that y = maxx, 1 forall x, y ∈ K.

(3) If (G,+,≤) is a totally ordered Abelian group, then (G∪−∞,max,+,≤),known as G-max-plus algebra, is a totally ordered positive semifield withthe property that for all y ≥ 0, there is an x ≥ −∞ such that y = maxx, 0.


(4) For nonnegative real numbers a, b, define a⊕h b = (a1/h + b1/h)h, where his a fixed positive real number. It is easy to see that Sh = (R≥0,⊕h, ·) is atotally ordered positive semifield with the property that for y ≥ 1, there isan x ≥ 0 such that y = x ⊕h 1 for all x, y. For more on the semiring Sh,one can refer to [24].

Remark 3.15. Let N be the set of natural numbers (positive integers) and a, b ∈ N.Then it is easy to check that if lcma, b and gcda, b denote the least common mul-tiple and greatest common divisor of the natural numbers a and b, then (N, lcm, gcd)is a pre-semiring [14, Example 6.5.8] and lcma, b · gcda, b = a · b [23, Theorem6.4]. This leads us to define a nontrivial modular function as follows:

Proposition 3.16. Define operations ⊕ and ⊙ on N as a ⊕ b = lcma, b anda ⊙ b = gcda, b. Then the function f : N −→ R defined as f(x) = log x ismodular, i.e.,

f(a⊕ b) + f(a⊙ b) = f(a) + f(b).

Let us recall that a semiring S is called simple if s+ 1 = 1 for any s ∈ S [12, p.4]. We first prove the following lemma for modular functions over simple semirings.After that, we are able to give another interesting nontrivial modular function.

Lemma 3.17. Let S be a simple semiring, G an Abelian group, and f : S −→ Ga modular function. Then f(xmyn) = f(xy) for any x, y ∈ S and m,n ∈ N.

Proof. Let x, y ∈ S and apply the modular equality for the elements y, yx. So,

f(y + yx) + f(y · yx) = f(y) + f(yx).

Also, y + yx = y(x + 1) = y. Therefore, f(xy2) = f(xy). Now by induction on mand n, it is easy to prove that f(xmyn) = f(xy).

Theorem 3.18. Let (G,+) be an Abelian group and D a Dedekind domain. Thenthe following statements hold:

(1) The semiring (Id(D),+, ·) is a simple semidomain.(2) If n ≥ 2 and p1, . . . , pn are distinct maximal ideals of D, then p1+p2 · · · pn =

D.(3) A modular function f : Id(D) −→ G can be characterized by the values of

f at (0), D, and all maximal ideals of D.

Proof. Let D be a Dedekind domain and a a nonzero and proper ideal of D. Thena = pα1

1 pα2

2 · · · pαk

k , where pi are maximal ideals of D, and αi > 0, k ≥ 2 are integers.Then by Lemma 3.17, we have

f(a) = f(pα1

1 pα2

2 · · · pαk

k ) = f(p1p2 · · · pk).

Now since pis are maximal, modularity of f gives the following equality:

f(D) + f(p1p2 · · · pk) = f(p1) + f(p2p3 · · · pk).

Applying modular property for the ideals p2 and p3p4 · · · pk, we get

2f(D) + f(p1p2 · · · pk) = f(p1) + (f(D) + f(p2p3 · · · pk))

= f(p1) + f(p2) + f(p3 · · · pk).

Continuing this process, we can write

f(a) = f(p1p2 · · · pk) = f(p1) + f(p2) + · · ·+ f(pk)− (k − 1)f(D).(3.3)


Our claim is that if the values of f at (0), D, and all maximal ideals of D aregiven, we can define f for an arbitrary nonzero proper ideal a = pα1

1 pα2

2 · · · pαk

k

(αi > 0 and k ≥ 1) of D as follows:

f(a) = f(p1) + f(p2) + · · ·+ f(pk)− (k − 1)f(D).(3.4)

and then f becomes modular.In order to show the trueness of our claim, we distinguish four cases for the

following formula:

f(a+ b) + f(a · b) = f(a) + f(b).(3.5)

(1) If one of the ideals a and b is the zero ideal, then it is straightforward thatequality (3.5) holds.

(2) Also, if one of the ideals a and b is the whole domain D, again it is straight-forward that equality (3.5) holds.

(3) By Proposition 6.8 in [19], if the ideals a = pα1

1 pα2

2 · · · pαk

k and b = qβ1

1 qβ2

2 · · · qβl

l

have no common maximal factor in their decompositions, then the ideals aand b are comaximal, i.e.,

a+ b = D,

and we have the following equalities:

f(a+ b) = f(D),

f(a) = f(p1) + f(p2) + · · ·+ f(pk)− (k − 1)f(D),

f(b) = f(q1) + f(q2) + · · ·+ f(ql)− (l − 1)f(D),

f(ab) = f(p1) + f(p2) + · · ·+ f(pk) + f(q1) + f(q2) + · · ·+ f(ql)− (k+ l− 1)f(D),

which show that equality (3.5) holds in this case as well.

(4) By Proposition 6.8 in [19], if the ideals a = pα1

1 pα2

2 · · · pαk

k and b = qβ1

1 qβ2

2 · · · qβl

l

have common maximal factors in their decompositions and those commonmaximal factors are the ideals r1, r2, . . . , rt, then a + b = r

γ1

1 rγ2

2 · · · rγt

t andthe following equalities satisfy:

f(a+ b) = f(r1) + f(r2) + · · ·+ f(rt)− (t− 1)f(D),

f(a) = f(p1) + f(p2) + · · ·+ f(pk)− (k − 1)f(D),

f(b) = f(q1) + f(q2) + · · ·+ f(ql)− (l − 1)f(D),

f(ab) = [k∑

i=1

f(pi) +l∑

j=1

f(qj)−t∑

s=1

f(rt)]− (k + l − t− 1)f(D),


Corollary 3.19. Let (G,+) be an Abelian group and f be a function from N0 intoG with the following property:

f(lcma, b) + f(ab) = f(a) + f(b).

Then f can be characterized by the values of f at numbers 0, 1, and all primenumbers of N.

Proof. The semiring (N0, lcm, ·) is isomorphic to the semiring (Id(Z),+, ·) and Z isa Dedekind domain.


Example 3.20. Let D be a Dedekind domain. Define a function f : D −→ Z asfollows:

(1) f((0)) = f(D) = 0.(2) If a = mα1

1 · · ·mαk

k , where m1, . . . ,mk are distinct maximal ideals of D, thenf(a) = k.

By Theorem 3.18, f is modular and one can interpret f as a function that countsthe number of maximal factors in the decomposition of an arbitrary ideal of D.

Now we go further to give other important examples for nontrivial modularfunctions in different branches of mathematics:

Example 3.21. Some examples for modular functions in different branches ofmathematics:

(1) Let Ω be a finite set and 2Ω the set of all subsets of Ω. The size function| · | : 2Ω −→ N0 is obviously a modular function. On the other hand,according to classical probability, the probability of an event A ⊆ Ω isdefined by P (A) = |A|/|Ω|. So, P (A ∪ B) + P (A ∩ B) = P (A) + P (B)for any A,B ⊆ Ω. This means that P : 2Ω −→ [0,+∞) is also a modularfunction [27]. It is clear that (2Ω,∪,∩, ∅,Ω,′ ) is a Boolean algebra andtherefore a semiring.

(2) Let V be a finite dimensional inner-product vector space. If we denote theset of all subspaces of V by Sub(V ), then the function dim : Sub(V ) −→ N0

is a modular function, since by Proposition 5.16 in [13], we have

dim(W1 +W2) + dim(W1 ∩W2) = dim(W1) + dim(W2).

It is clear that (Sub(V ),+,∩, (0), V,⊥ ) is a Boolean algebra and thereforea semiring.

(3) Let M be an R-module of finite length and let K and N be R-submodulesof M . Then c(K +N) + c(K ∩N) = c(K) + c(N), where c(L) is the com-position length of a module [2, Corollary 11.5]. Now, if we consider M tobe a distributive module [6] of finite length, then Sub(M) is a bounded dis-tributive lattice and therefore, a semiring. For an example of a distributivemodule of finite length, refer to [7].

(4) Let D be an integral domain. Then D is a Prufer domain if and only if(I + J)(I ∩ J) = IJ for all ideals I, J of D. On the other hand, if D is aPrufer domain, then (Id(D),+,∩) is a semiring [22, Theorem 6.6], where byId(D), we mean the set of all ideals of D. Now it is clear that the functionf from the semiring (Id(D),+,∩) to the commutative semigroup (Id(D), ·)defined by f(I) = I is a modular function.

(5) Let (G,+) be a partially-ordered Abelian group. It is easy to see that a∨ bexists if and only if a ∧ b exists in G and in this case, by Theorem 15.2 in[10],

a ∨ b+ a ∧ b = a+ b.

Therefore, in such a case, the function f : (G,∨,∧) −→ (G,+) defined byf(x) = x is a modular function. Now, if (G,∨,∧) is also a distributivelattice, f is a modular function over the pre-semiring G.

(6) Let L be a bounded distributive lattice of finite length. Then the functionheight : L −→ N0 is a modular function, since

height(a ∨ b) + height(a ∧ b) = height(a) + height(b)


for all a, b ∈ L [15, Corollary 376].(7) Consider the Peano-Jordan content c (volume, area or length) over an Eu-

clidian space En. Define FC(En) to be the set of all subsets A ⊆ En suchthat c(A) < ∞. Then (FC(En),∪,∩, ∅) is a hemiring and c is a modularfunction over FC(En) [26, Theorem 22].

Now, we give a pre-semiring version of the so-called inclusion-exclusion theoremin probability theory. Note that the set-theoretic version of the following theoremis due to Poincare [4, Exercise 1.12.52.]. It is also a good point to mention that onecan see the number theory version of the inclusion-exclusion theorem in Theorems7.3 and 7.4 in [23]. Also, for a multiplicative ideal theory version of this theoremrefer to the recent paper [1] by D. D. Anderson et al. on GCD and LCM-likeidentities for ideals in commutative rings.

Theorem 3.22 (Inclusion-Exclusion Theorem). Let S be a multiplicatively idem-potent pre-semiring and T a commutative semigroup. If m : S −→ T is a modularfunction, then for any elements s1, s2, . . . , sn in S, the following equality holds:

m

(

n∑

i=1

si

)

+∑

i<j

m(sisj) +∑

i<j<k<l


n∑

i=1

m(si) +∑

i<j<k



Proof. The proof is by mathematical induction. Since m is modular, the Poincare’sFormula holds for k = 2. Now let Poincare’s Formula hold for k = n and takes1, . . . , sn, sn+1 ∈ S. Since m is modular, we have

m(

n∑

i=1

si + sn+1

)

+m(

(

n∑

i=1

si) · sn+1

)

=

n∑

i=1

m(si) +m(sn+1).

Now by applying Poincare’s Formula for the n elements s1sn+1, s2sn+1, . . . , snsn+1,we have

m(

n∑

i=1

sisn+1

)

+∑

i<j

m(sisn+1sjsn+1)+∑

i<j<k<l

m(sisn+1sjsn+1sksn+1slsn+1)+· · · =

n∑

i=1

m(sisn+1) +∑

i<j<k

m(sisn+1sjsn+1sksn+1) + · · · .

By using that S is multiplicatively idempotent, the above equality can be writtenas follows:

m(

n∑

i=1

sisn+1

)

+∑

i<j

m(sisjsn+1) +∑

i<j<k<l

m(sisjskslsn+1) + · · · =

n∑

i=1

m(sisn+1) +∑

i<j<k

m(sisjsksn+1) + · · · .


Finally, by adding m(∑n

i=1 si + sn+1) to both sides of the recent equality, we getthe Poincare’s Formula for the n + 1 elements s1, s2, . . . , sn+1 and the proof iscomplete.

Proposition 3.23. Let S be a multiplicatively idempotent pre-semiring, R a ring,and f : S −→ R a modular function. For any n > 1, if s1, s2, . . . , sn in S areindependent, then s1 + s2 + · · ·+ sn−1 and sn are also independent.

Proof. Since S is multiplicatively idempotent, by Theorem 3.22, we have:

f(

(s1 + · · ·+ sn−1)sn)

= f(s1sn + · · ·+ sn−1sn) =

n−1∑

i=1

f(sisn)−∑

1≤i<j<n

f(sisjsn)+∑

1≤i<j<k<n

f(sisjsksn)−· · ·+(−1)nf(s1s2 · · · sn) =

[

n−1∑

i=1

f(si)−∑

1≤i<j<n

f(sisj)+∑

1≤i<j<k<n

f(sisjsk)−· · ·+(−1)nf(s1s2 · · · sn−1)]

f(sn) =

f(s1 + · · ·+ sn−1)f(sn),


In the following theorem and its corollaries, we show when a finitely additivefunction becomes modular:

Theorem 3.24. Let S be a zerosumfree semiring such that 1 + 1 ∈ comp(S), Ta commutative semigroup, and f : S −→ T a finitely additive function. Thenf |comp(S): comp(S) −→ T is modular.

Proof. Let s, t ∈ comp(S). Therefore by Proposition 1.4 and Theorem 2.4, we havef(s + t) = f(s ⊔ t) = f(s + s⊥t) = f(s) + f(s⊥t). Finally, by Proposition 1.4,f(s+ t)+f(st) = f(s)+f(s⊥t)+f(st) = f(s)+f(t) and the proof is complete.

By Theorem 2.4, it is clear that if each element of a semiring S is complemented,then S is a Boolean algebra. Therefore, it is obvious that we have the followingresult:

Corollary 3.25. Let S be a Boolean algebra and T a commutative and cancellativemonoid. Then the following statements for a function f : S −→ T are equivalent:

(1) f is finitely additive,(2) f is modular and f(0) = 0.

Note that if L is a Stone algebra, then Skel(L) is a Boolean algebra [15, Theorem214]. Therefore, we have the following:

Corollary 3.26. Let L be a Stone algebra and T a commutative and cancelationmonoid. If f : L −→ T is finitely additive, then f |Skel(S): Skel(S) −→ T ismodular.

Proposition 3.27. Let S be a zerosumfree semiring such that 1 + 1 ∈ comp(S),T a commutative semigroup, and f : S −→ T a finitely additive function. For anys1, s2, . . . , sn ∈ comp(S), we have the following:

m(

n∑

i=1

si)

+∑

i<j

m(sisj) +∑

i<j<k<l



n∑

i=1

m(si) +∑

i<j<k



Proof. Let S be a zerosumfree semiring such that 1 + 1 ∈ comp(S). So by The-orem 2.4, comp(S) is a Boolean algebra. This means that comp(S) is a multi-plicatively idempotent semiring. On the other hand, by Theorem 3.24, m |comp(S):comp(S) −→ T is a modular function. Hence, by Theorem 3.22, Poincare’s Formulaholds for complemented elements of S and this completes the proof.

Acknowledgment

The first named author is supported by Department of Engineering Science atGolpayegan University of Technology. The authors wish to thank Sepanta Asadifor informing them about a point related to Proposition 3.7. They are also gratefulfor the useful comments of the anonymous referee.

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2003.

Peyman Nasehpour, Department of Engineering Science, Golpayegan University of

Technology, Golpayegan, Iran

E-mail address: [email protected], [email protected]

Amir Hossein Parvardi, Department of Mathematics, University of British Columbia,

Vancouver, BC, Canada V6T 1Z2

E-mail address: [email protected]

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FINITELY ADDITIVE, MODULAR, AND PROBABILITY FUNCTIONS …